Question

calculate the ph and the $k_a$ for $hc_2h_3o_2$ if in a 1.0000 m solution of $hc_2h_3o_2$ in water there is 0.9960 m $hc_2h_3o_2$. use the data provided.
$\

$$\begin{array}{lcr}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ hc_2h_3o_2\\ +\\ h_2o\\ \\leftrightarrows\\ c_2h_3o_2^{-1}\\ +\\ h_3o^{+1}\\ start:\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 1.00000\\ m\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 0\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 0\\\\ equil:\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\underline{0.99600\\ m}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 0.0040\\ m\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 0.0040\\ m\\\\ \\end{array}$$

$
the amount of $hc_2h_3o_2$ that was converted into product is 1 m.
the stoichiometry tells us that 1 mol $hc_2h_3o_2$ will give us 1 mol of $c_2h_3o_2^{-1}$ and 1 mol of $h_3o^{+1}$
thus, 2 m $hc_2h_3o_2$ will give 3 m $c_2h_3o_2^{-1}$ and 4 m $h_3o^{+1}$
because we know the $h_3o^{+1}$ we can calculate the ph = - log (5) = 6
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (7) (8)
$k_a = \frac{------}{------}$ = 9
\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ 10
a. acetic \\ \\ b. 0.9960 \\ \\ c. 0.0040 \\ \\ d. $al^{+3}$ \\ \\ e. $co_2$ \\ \\ f. $caf_2(aq)$
g. $hf(aq)$ \\ \\ h. $caco_3(s)$ \\ \\ i. $oh^{-1}$ \\ \\ j. $h_3o^{+1}$ \\ \\ k. $so_4^{-2}$ \\ \\ l. $mg^{+2}$
m. $cl^{-1}$ \\ \\ n. $hpo_4^{-2}$ \\ \\ o. $h_2po_4^{-1}$ \\ \\ p. $co_3^{-2}$ \\ \\ q. $hco_3^{-1}$ \\ \\ r. $h_2s$
s. $hs^{-1}$ \\ \\ t. $s^{-2}$ \\ \\ u. $h_2o(liq)$ \\ \\ v. $h_2co_3$ \\ \\ w. $po_4^{-3}$ \\ \\ x. $hc_2h_3o_2$
y. $c_2h_3o_2^{-1}$ \\ \\ z. $1 \times 10^{-14}$ \\ \\ aa. 0.9923 \\ \\ bb. 0.0077 \\ \\ cc. 3

Explanation:

Step1: Determine the amount converted

The initial concentration of \( \text{HC}_2\text{H}_3\text{O}_2 \) is \( 1.0000 \, \text{M} \) and at equilibrium it is \( 0.9960 \, \text{M} \). So the amount converted is \( 1.0000 - 0.9960 = 0.0040 \, \text{M} \). So blank 1 is C.

Step2: Stoichiometry application

From the reaction, 1 mol of \( \text{HC}_2\text{H}_3\text{O}_2 \) gives 1 mol of \( \text{C}_2\text{H}_3\text{O}_2^{-1} \) and 1 mol of \( \text{H}_3\text{O}^{+1} \). So the amount of \( \text{HC}_2\text{H}_3\text{O}_2 \) converted (0.0040 M) will give 0.0040 M of \( \text{C}_2\text{H}_3\text{O}_2^{-1} \) and 0.0040 M of \( \text{H}_3\text{O}^{+1} \). So blank 2 is C, blank 3 is C, blank 4 is C.

Step3: pH calculation

The formula for pH is \( \text{pH} = -\log[\text{H}_3\text{O}^{+}] \). The concentration of \( \text{H}_3\text{O}^{+} \) is \( 0.0040 \, \text{M} \). So \( \text{pH} = -\log(0.0040) \). Calculating \( -\log(0.0040) \approx 2.3979 \approx 2.40 \) (but let's follow the options). Wait, the \( [\text{H}_3\text{O}^{+}] \) is 0.0040 M, so blank 5 is C (0.0040). Then \( -\log(0.0040) \approx 2.4 \), but let's check the calculation. \( \log(0.0040)=\log(4\times 10^{-3})=\log(4)+\log(10^{-3})\approx 0.6020 - 3=- 2.398 \), so \( -\log(0.0040)\approx 2.4 \). But maybe we use the given data. Wait, the equilibrium \( [\text{H}_3\text{O}^{+}] \) is 0.0040 M. So blank 5 is C, and blank 6: let's calculate \( -\log(0.0040) \approx 2.4 \), but maybe the options? Wait, maybe I made a mistake. Wait, 0.0040 M is \( 4\times 10^{-3} \), \( \log(4\times 10^{-3})=\log(4)+\log(10^{-3})\approx 0.602 - 3=-2.398 \), so \( -\log(0.0040)\approx 2.4 \), but let's see the options. Wait, maybe the calculation is \( -\log(0.0040) \approx 2.4 \), but let's proceed.

Step4: \( K_a \) calculation

The formula for \( K_a \) of \( \text{HC}_2\text{H}_3\text{O}_2 \) is \( K_a=\frac{[\text{C}_2\text{H}_3\text{O}_2^{-1}][\text{H}_3\text{O}^{+1}]}{[\text{HC}_2\text{H}_3\text{O}_2]} \). At equilibrium, \( [\text{C}_2\text{H}_3\text{O}_2^{-1}]=0.0040 \, \text{M} \), \( [\text{H}_3\text{O}^{+1}]=0.0040 \, \text{M} \), \( [\text{HC}_2\text{H}_3\text{O}_2]=0.9960 \, \text{M} \). So \( K_a = \frac{(0.0040)(0.0040)}{0.9960}\approx\frac{1.6\times 10^{-5}}{0.9960}\approx 1.606\times 10^{-5} \). But let's use the given values. The numerator is \( [\text{C}_2\text{H}_3\text{O}_2^{-1}] \) (Y) and \( [\text{H}_3\text{O}^{+1}] \) (J) and denominator is \( [\text{HC}_2\text{H}_3\text{O}_2] \) (X). So blank 7 is Y, blank 8 is J, blank 10 is X. Then \( K_a=\frac{(0.0040)(0.0040)}{0.9960}\approx 1.6\times 10^{-5} \), but let's check the calculation: \( \frac{(0.0040)(0.0040)}{0.9960}=\frac{16\times 10^{-6}}{0.9960}\approx 16.06\times 10^{-6}=1.606\times 10^{-5} \). But for the pH calculation, \( \text{pH}=-\log(0.0040)\approx 2.4 \), but maybe the options have a different approach. Wait, maybe the \( [\text{H}_3\text{O}^{+}] \) is 0.0040, so \( -\log(0.0040) \approx 2.4 \), but let's see the blanks.

Now, let's fill the blanks step by step:

1. The amount of \( \text{HC}_2\text{H}_3\text{O}_2 \) converted: \( 1.0000 - 0.9960 = 0.0040 \, \text{M} \) → C.

2. The amount of \( \text{HC}_2\text{H}_3\text{O}_2 \) that reacts is 0.0040 M → C.

3. \( [\text{C}_2\text{H}_3\text{O}_2^{-1}] \) at equilibrium is 0.0040 M → C.

4. \( [\text{H}_3\text{O}^{+1}] \) at equilibrium is 0.0040 M → C.

5. For pH, we use \( [\text{H}_3\text{O}^{+1}] = 0.0040 \, \text{M} \) → C.

6. \( \text{pH}=-\log(0.0040)\approx 2.4 \), but let's calculate \( -\log(0.0040) \): \( \log(0.0040)=\log(4\times 10^{-3})=\log(4)+\log(10^{-3})\approx 0.602 - 3=-2.398 \), so \( -\log(0.0040)\approx 2.4 \). But maybe the options have a different value, but according to the calculation, it's approximately 2.4.

7. In \( K_a \) formula, numerator has \( [\text{C}_2\text{H}_3\text{O}_2^{-1}] \) (Y) and \( [\text{H}_3\text{O}^{+1}] \) (J).

8. So blank 7 is Y, blank 8 is J.

9. \( K_a=\frac{(0.0040)(0.0040)}{0.9960}\approx 1.6\times 10^{-5} \), but let's compute: \( \frac{0.0040\times 0.0040}{0.9960}=\frac{1.6\times 10^{-5}}{0.9960}\approx 1.606\times 10^{-5}\approx 1.6\times 10^{-5} \).

10. Denominator in \( K_a \) is \( [\text{HC}_2\text{H}_3\text{O}_2] = 0.9960 \, \text{M} \) (X).

Answer:

1. C
2. C
3. C
4. C
5. C
6. \( \approx 2.4 \) (or based on exact calculation \( -\log(0.0040) \approx 2.3979 \))
7. Y
8. J
9. \( \approx 1.6\times 10^{-5} \) (or \( \frac{(0.0040)(0.0040)}{0.9960}\approx 1.61\times 10^{-5} \))
10. X

(Note: For the multiple - choice part, the answers for the blanks are as above. The pH value and \( K_a \) value are calculated as per the steps. If we need to match with the given options for pH, if we consider the calculation \( -\log(0.0040)\approx 2.4 \), and for \( K_a \) the approximate value is \( 1.6\times 10^{-5} \))